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Let $c,d$ be natural numbers of same parity (both odd or both even) and $\sigma$ be sum of divisors function. Is it known whether or under what conditions $\sigma (c^{2})$=$\sigma (d^{2})$ ? I am guessing that it can happen but unsure if certain things must hold

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  • $\begingroup$ Have you seen this? $\endgroup$ – Don Thousand Dec 24 '18 at 19:19
  • $\begingroup$ Thank you,I don't know if that helps me $\endgroup$ – argamon Dec 24 '18 at 19:21
  • $\begingroup$ @argamon I'd start with a simpler problem - when $c, d$ are odd primes, it simplifies to solving :$$1+c+c^2 =1+d+d^2$$ $\endgroup$ – AgentS Dec 24 '18 at 20:08
  • $\begingroup$ I bet you had tried that, not that interesting.. $\endgroup$ – AgentS Dec 24 '18 at 20:12
  • $\begingroup$ Haha, it's just more useful if I know in general but I greatly appreciate your help $\endgroup$ – argamon Dec 24 '18 at 20:13
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It looks to me like $\sigma(627^2)=\sigma(749^2)$.

Feel free to check my work. (Or Python's work!)

And here's an even-even pair: $\sigma(740^2)=\sigma(878^2)$.

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As pointed out in the comments, starting from the simple example $$\sigma (4^2)=\sigma (5^2)$$ we can generate infinitely many by multiplying by a factor prime to $10$. Thus, $$\sigma(12^2)=(1+2+2^2+2^3+2^4)\times (1+3+3^2)=31\times 13=403$$ $$\sigma(15^2)=(1+3+9)\times (1+5+25)=13\times 31=403$$

and so on.

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  • $\begingroup$ That's very helpful I wonder if there exist any with the same parity i.e. both odd or both even $\endgroup$ – argamon Dec 24 '18 at 19:45
  • $\begingroup$ I suggest doing a search. I also found $\sigma(76^2)=\sigma(95^2)$ but I didn't search very far. Of course, if two odd numbers satisfied this you could multiply by $4$ to get an example with two even numbers. Of course both of my examples use the fact that $1+2+2^2+2^3+2^4=1+5+5^2$. And it's easy to make more examples using that. I'd first want to see if there were examples without that. $\endgroup$ – lulu Dec 24 '18 at 19:48
  • $\begingroup$ In fact, the sum of divisors function is multiplicative, so you can multiply both of them by the square of any number that does not have $2,3,5$ as a factor and get another pair. $\endgroup$ – Ross Millikan Dec 24 '18 at 20:22
  • $\begingroup$ You have $\sigma(4^2)=\sigma(5^2)=31$ and now you can multiply by the square of any number without factors of $2,5$. The answer is $3^2$ and $76^2,95^2$ is $19^2$ $\endgroup$ – Ross Millikan Dec 24 '18 at 20:44
  • $\begingroup$ @RossMillikan Yes, both my examples are simple consequences of $\sigma(4^2)=\sigma (5^2). I'll edit to point that out, $\endgroup$ – lulu Dec 24 '18 at 20:52

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